This has now been cross posted to MO.

Let $F$ be a subset of $\mathbb{R}$ and let $S_F$ denote the set of values which satisfy some generalized polynomial whose exponents and coefficients are drawn from $F$. That is, we let $S_F$ denote $$\bigg \{x \in \mathbb{R}: 0=\sum_{i=1}^n{a_i x^{e_i}}: e_i \in F \text{ distinct}, a_i\in F \text{ non-zero}, n\in \mathbb{N} \bigg \}$$

Then $S_{\mathbb{\mathbb{Q}}}$ is the set of algebraic real numbers and we start to see the beginnings of a chain:

$ \mathbb{Q} \subsetneq S_\mathbb{Q} \subsetneq S_{S_\mathbb{Q}} $

Main Question

Does this chain continue forever? That is, we let $A_0= \mathbb{Q}$ and let $A_{n+1}=S_{A_{n}}$. Is it the case that $A_n \subsetneq A_{n+1}$ for all $n\in\mathbb{N}$?

Other curiosities:

Is $A_i$ always a field? Perhaps, the argument is analogous to this. Or maybe this is just the case in a more general setting: Is it the case that $F \subset \mathbb{R}$, a field implies that $S_F$ is a field?

Is it possible to see that $e\notin \cup A_i$? Perhaps this is just a tweaking of LW Theorem.

  • 3
    $\begingroup$ Every set in the chain is countable, so you certainly don't hit $\mathbb{R}$ at any point. $\endgroup$ – Patrick Stevens Nov 26 '18 at 19:19
  • $\begingroup$ How do you propose to define $x^\alpha$ if $\alpha \notin \mathbb{Q}$? $\endgroup$ – Hans Engler Nov 26 '18 at 19:24
  • 1
    $\begingroup$ I'm still somewhat concerned about the definition of $x^\alpha$ when $x<0$ and $\alpha$ is not an integer. But, there is a cool question in here (possibly a very difficult one). $\endgroup$ – Jyrki Lahtonen Dec 3 '18 at 4:31
  • 1
    $\begingroup$ Mason, I had some simple worries in my mind. Like is $S_F$ necessarily closed under negation? $\endgroup$ – Jyrki Lahtonen Dec 3 '18 at 14:06
  • 2
    $\begingroup$ I don’t think we can rule out $e\in A_3$. My impression is that we know almost nothing about the transcendence of numbers like $3^\sqrt6-4^\sqrt3-1$. $\endgroup$ – Matt F. Dec 6 '18 at 22:35

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